Comments on: Thanksgiving

By: Dave A. (Dr. EBeam)

Dave A. (Dr. EBeam) — Mon, 15 Dec 2008 00:55:36 +0000

Regarding the “force” due to the exclusion principle.
Someone once explained it to me (I think it was either Kurt Gottfried or Victor Weiskopf) that the exclusion principle creates a pressure, which keeps things from passing right through each other. That pressure can be thought of as due to the kinetic energy of the electrons. As one tries to compress the electrons, their kinetic energy increases. Similarly, trying to put more electrons (or atoms with electrons) in the same space, will increase the kinetic energy of the electrons.

The kinetic energy increases because the energy levels depend on the volume the electrons are allowed to occupy. A smaller volume (or more electrons in the same volume) will increase the total energy of the system. For some reason, we like to think of a force as arising from a potential energy, but we have no problem with a pressure being due to the kinetic energy of a gas. The exclusion principle is just the pressure due to the kinetic energy of the electrons–which increases when they are compressed. (It’s a wave-function thing, not dependent on temperature.)

-Dave

By: Pat Dennis

Pat Dennis — Mon, 01 Dec 2008 17:25:44 +0000

Matt — Nothing here specifically for iPhone, but try these:

http://www.particlezoo.net/
http://www.youtube.com/watch?v=j50ZssEojtM
… and especially: http://www.particleadventure.org/

By: Matt

Matt — Mon, 01 Dec 2008 15:07:07 +0000

Thanks, Sean, that helped clear up some confusion I’ve always had over force particles. Now what I’d love is if someone would put the bestiary into an iPhone app, preferably with little visualizations for each particle to show my kids when they ask me.

By: Carl Brannen

Carl Brannen — Mon, 01 Dec 2008 03:53:11 +0000

Neil,

You can use density matrices in something like 5 very much different ways. One of them is as you’ve described. But you can also take the pure states that are used in state vectors and convert them into “pure density matrices”. These are exactly equivalent to the usual state vectors in terms of information content, except that state vectors have arbitrary complex phases that have no physical interpretation. (The physical uses of complex phase are shared with between density matrices, otherwise they wouldn’t work.)

“So I wonder what physical significance there would have to be to a DM formulation, considering it is “a way of looking at things” more than an actual situation of thing.”

If you’re interested in the actual thing (or ontology), rather than mathematical ways of describing activity, then you’re probably interested in Bohmian mechanics. I’ve linked to a paper that discusses density matrices with respect to Bohmian mechanics which was the first reference I saw on arXiv in 30 seconds, but there are a lot of more complete articles. It’s an active field.

From the density matrix point of view, state vectors are a kluge that is done so that a quantum state can be converted from its natural bilinear condition, to an unnatural, but mathematically convenient linear one. The mathematical convenience is “linear superposition”, which is a principle that allows you to describe a large number of quantum states from taking linear combinations of other states, for example, the usual basis states in spin-1/2 of spin up and spin down.

Linear superposition (as opposed to interference) does not translate into a literal experiment that can be performed in the lab. That would be like taking, for example, combining a spin up electron with a spin down electron and expecting to get an electron oriented with spin in the x direction. You will not get this as a result. Instead, you will get two electrons and to model them you will require a more complicated state vector, i.e. ++, +-, -+, — as basis. Of course interference works fine in density matrix calculations just like with state vectors.

The bilinearity of density matrices can be thought of as describing a quantum state in terms of how it acts as an operator. The pure density matrices are projection operators. This makes density matrices, rather than state vectors, the natural way to define the “Consistent Histories” interpretation of quantum mechanics (which generalizes QM to quantum cosmology). Read the article on consistent histories at Wikipedia, or do a search for arXiv articles.

So density matrices have an interpretation of quantum mechanics that is devoted to their use and avoids the state vector formalism. At the very least this makes density matrices the equal of state vectors but for those who prefer quantum mechanics that naturally fits into quantum cosmology, density matrices are superior. And their sign does not change when you swap particles in them.

Mathematically, this comes about because a density matrix is made from copies of the same state vector, a bra and a ket. When you swap two particles, you get a minus sign for the bra and a minus sign for the ket. Since (-1)(-1) = +1, the density matrix is unchanged.

By: Neil B

Neil B — Sun, 30 Nov 2008 22:16:11 +0000

Carl Brannen, isn’t it so that the “density matrix” is a sort of kludge representing uncertain knowledge of a wave function, it isn’t considered to be a “true” representation of a given actual WF. For example, I may have a source that emits linear polarized light, and I don’t know the details of how they are made (maybe, I don’t have a distinct linear filter available, maybe my assistant keeps turning it around, etc.) But must I think that “real waves” can literally not have a definite polarization (unless entangled)? So I wonder what physical significance there would have to be to a DM formulation, considering it is “a way of looking at things” more than an actual situation of thing.

By: Pat Dennis

Pat Dennis — Sun, 30 Nov 2008 18:54:43 +0000

Sean, this post was crystal-clear, as was the one on Hubble’s Law…. but I’d sure like to have seen something a little more accessible on the Lagrangian of the Standard Model of particle physics. Four semesters of college physics, and calculus through elementary differential equations, didn’t – in my case, at least – even come close to preparing me for that one!!

By: Carl Brannen

Carl Brannen — Sun, 30 Nov 2008 18:01:21 +0000

“The answer is, Fermi statistics: when you take two identical fermions and switch them with each other, you get minus the state you started with.”

This is true only for the state vector formalism of quantum mechanics. In density matrix formalism, the arbitrary complex phases are eliminated and exchanging two particles leaves one with the same density matrix state.

Since density matrices and state vectors give identical results in quantum mechanics, it is not possible to say which one is fundamental. Consequently, it cannot be said that swapping fermions changes a state.

From the density matrix point of view, a state vector is a sort of square root of a density matrix. It is not surprising that one must supply a phase (in analogy with the sign that arises with a square root). But if the fundamental nature is the density matrix (which can represent states that state vectors cannot and therefore are more general), then the minus sign that shows up when fermions are swapped is just an accident of mathematics, not a fundamental part of the physics. Instead, the difference between fermions and bosons is that the occupation numbers for fermions can only be 0 or 1 while bosons can count higher.

Google “nine formulations of quantum mechanics” for more info.

By: Ars Mathematica » Blog Archive » Thanks for Spin-Statistics Theorem

Ars Mathematica » Blog Archive » Thanks for Spin-Statistics Theorem — Sun, 30 Nov 2008 04:20:55 +0000

[...] little tradition where for each Thanksgiving Day they pick a physics result to be thankful for. This year they pick the spin-statistics theorem, which explains why elementary particles with half-integer [...]

By: Chris W.

Chris W. — Sun, 30 Nov 2008 03:02:04 +0000

As a matter of fact, Broyles did post a preprint in 2002 (last revised in Feb 2003). Its subject is also quite interesting: One Loop Vacuum Polarization without Infinities (hep-th/0207069)

A technique for avoiding infinite integrals in the calculation of the one-loop diagram contribution to the vacuum polarization component of an atomic energy level is presented. This makes renormalization unnecessary. Infinite integrals do not occur because, as it is shown, no delta functions are required for the Green's functions. Thus there are none to overlap. This procedure is shown to produce the same formula as the one obtained by dimensional renormalization.

(But I digress...)

By: Fermi-Walker Public Transport

Fermi-Walker Public Transport — Sat, 29 Nov 2008 15:53:03 +0000

The Arthur Broyles paper came out in 1999,
here is a link: arxiv.org/abs/quant-ph/9906046

By: Jimbo

Jimbo — Sat, 29 Nov 2008 08:18:29 +0000

For Robert,
Look up the 2002 paper by Arthur Broyles on the arxiv. He recites some of
the history of the SST, from Feynman’s bemoaning that, other than Pauli’s 1940 QFT derivation, no simple, intuitive derivation existed. Broyles provided such a simple derivation from the properties of wave functions alone, sans QFT. It was promptly attacked by Duck & Sudarshan, but to my knowledge, is the only such `simple’ SST derivation to date. Why it has not made it to textbook status, I do not know.

By: Amiya Sarkar

Amiya Sarkar — Sat, 29 Nov 2008 06:45:39 +0000

Great! It reminded me of my passion in particle physics. The passion is still in my heart but it got in a Boson state piling up with other disciplines of science. I think, may be wistfully, that all these fermions and bosons are mere wave functions which may somehow be explained in ‘string theory’ terms.
The bloody mayhem in India, Bose’s homeland, is very unfortunate. I prefer to call them ‘stupid cowards’ and not ‘terrorists’.
Thanks again for the mindboggling post, in no ‘uncertain’ terms!

By: harshpotatoes

harshpotatoes — Fri, 28 Nov 2008 19:47:19 +0000

Ah, so going back to my derivation of the pauli exclusion. More reasoning behind again comes back to the parity operator. If P is the parity operator, f(x,y) is the wavefunction, then we note that P^2(f(x,y)) = f(x,y). Apply the parity operator twice, which switches the particles twice, leaving them exactly where the were in the beginning. Assuming that the parity operator is hermitian, meaning it is observable, means that its eigenvalues can only be real, meaning there are only two possibilities: 1 or -1. (squareroot of 1 is either 1 or -1). which leads to the conclusion that in this simplified case of two particles in one dimension there are two types of wavefunctions, one we will call bosons and the other fermions.

So i guess coming back to this possibility of anyons, either the parity operator is not an observable for them, or they can’t exist in one dimension.

however, even this simple proof, while not a generalized proof for any dimension with any number of particles, it does show that there is a type of wavefunction which has zero probability when particles are at the same location.

By: MZ

MZ — Fri, 28 Nov 2008 18:40:28 +0000

This was great. I wish you would write a whole series like this. “An Intuitive Explanation of the Universe.”

By: Chris W.

Chris W. — Fri, 28 Nov 2008 17:07:50 +0000

(Off-topic...) Speaking of fermions and forces, I look forward to a bit of coverage on CV of the article that appeared in Science last week (21 Nov 2008, Vol. 322. no. 5905, pp. 1224 - 1227): Ab Initio Determination of Light Hadron Masses

We present a full ab initio calculation of the masses of protons, neutrons, and other light hadrons, using lattice quantum chromodynamics. ....... .... .... Our results completely agree with experimental observations and represent a quantitative confirmation of this aspect of the Standard Model with fully controlled uncertainties.

By: Lawrence Crowell

Lawrence Crowell — Fri, 28 Nov 2008 15:02:54 +0000

The exclusion principle is a topological aspect of QFT. Given the Fermionic field Y then Y^2 = 0, which is topologically the same as d^2 = 0. Topology further comes in with BRST, for SUSY operators Q or Q^* we have Q^2 = 0 and states |Y> are determined by ker(Q)/im(Q) so that a physical state is NOT determined by |Y> = Q|X>.

The exclusion principle is of course not a force, for a force is something which is determined locally or by some local gauge/phase change. It just means that for two fermions, say electrons in an atom with eigenvalues E,l,s, etc cannot occupy the same state (Y^2 = 0), which is a topological obstruction that prevents any “force” from changing an energy state &E = F*&x so two electrons occupy the same quantum state.

We can be thankful for this, for it makes chemistry possible. All that tryptophan would not be possible otherwise.

Lawrence B. Crowell

By: Callum M.

Callum M. — Fri, 28 Nov 2008 14:04:00 +0000

@Neil B:

The “Bosun” is a corruption of the term “Boatswain”, who is the chief maintenance officer aboard a ship. He’s also the one who “pipes” the captain and other ranking officers aboard.

Just a random spot of information there

By: Pieter Kok

Pieter Kok — Fri, 28 Nov 2008 08:04:05 +0000

The exclusion principle is not a force because it is a consequence of the structure of quantum mechanics, rather than a (set of) terms in the Lagrangian. Also, I do not like the “derivation” of the exclusion principle by harshpotatoes (no offense): you have shifted the problem to showing that wave functions are either symmetric or anti-symmetric. There is no intrinsic reason why this must be so (at least, not in quantum mechanics). For example, there is currently a lot of research going on about anyons, particles that do not have (half-) integer spin. Their wavefunctions have no particular symmetry.

By: miller

miller — Fri, 28 Nov 2008 01:29:23 +0000

I think degeneracy pressure is basically caused by the fact that if we add more and more electrons into the same amount of volume, they must occupy higher and higher energy states because the lower states have already been filled. Therefore, in absence of greater forces, it requires less energy for the electrons to remain spread out. The magnitude of this “force” is equal to the gradient in the energy required for those higher energy states.

By: Neil B

Neil B — Fri, 28 Nov 2008 00:54:21 +0000

Harshpotatoes, thanks and also your explanation of the essentially “semantic” reason for not calling degeneracy pressure “fundamental” is likely appropriate. However, the trouble remains that it is still “insinuated” in the framing of the issue that all forces between particles derive from the typical fundamental four: EM, grav, strong, weak. OK, call the exclusion force “quantum mechanical”, not mediated etc, but it is still ipso facto “a force.” See, I care about the proper use of language.
And BTW if exclusion is a matter of principle (?), what determines the magnitude of the force/s derived from it?

Also, food for strange thoughts: nuclei can have integer or half-integer spins (from the combination of effects despite each nucleon being a fermion.) This is basically the idea behind Einstein-Bose condensate (although put forth as the spin of “atoms”) but then if so fundamental, why the need to cool the stuff so much? (I mean, to get them to intermingle per se more so than to achieve a single quantum state – after all photons pack together without having to be in the same state – but pardon any middle-brow confusion.) Messiness caused by the associated electrons?

In any case, I don’t see much talk about distribution of fermionic v. bosonic nuclei in various stars at high density; pre-neutronium stage. Yet it seems it ought to affect the shrinkage and collapse of the star.